Goldstein_31_32_7.pdf

Homework 5: # 3.31, 3.32, 3.7a

Michael Good

Sept 27, 2004

3.7a Show that the angle of recoil of the target particle relative to the incident direction of the scattered particle is simply Φ =1₂(π−Θ).

Answer:

It helps to draw a figure for this problem. I don’t yet know how to do this in LA_{TEX, but I do know that in the center of mass frame both the particles}

momentum are equal.

m1v01=m2v20

Where the prime indicates the CM frame. If you take equation (3.2) Gold-stein, then its easy to understand the equation after (3.110) for the relationship of the relative speedv after the collision to the speed in the CM system.

v0₁= µ m1

v= m2 m1+m2

v

Here, v is the relative speed after the collision, but as Goldstein mentions because elastic collisions conserve kinetic energy, (I’m assuming this collision is elastic even though it wasn’t explicitly stated), we havev =v0, that is the

relative speed after collision is equal to the initial velocity of the first particle in the laboratory frame ( the target particle being stationary).

v₁0 = m2 m1+m2

v0

This equation works the same way forv20

v₂0 = m1 m1+m2

v0

From conservation of momentum, we know that the total momentum in the CM frame is equal to the incident(and thus total) momentum in the laboratory frame.

(m1+m2)vcm=m1v0

vcm=

m1

m1+m2

v0

This is the same asv20

v0₂=vcm

If we draw both frames in the same diagram, we can see an isosceles triangle where the two equal sides arev₂0 andvcm.

Φ + Φ + Θ =π

Φ = 1

2(π−Θ)

3.31 Examine the scattering produced by a repulsive central force f +kr−3_.

Show that the differential cross section is given by

σ(Θ)dΘ = k 2E

(1−x)dx x2_{(2− −x)}2_sinπx

wherexis the ratio of Θ/π andEis the energy.

Answer:

The differential cross section is given by Goldstein (3.93):

σ(Θ) = s sin Θ

ds dΘ

We must solve fors, andds/dΘ. Lets solve for Θ(s) first, take its derivative with respect tos, and invert it to findds/dΘ. We can solve for Θ(s) by using Goldstein (3.96):

Θ(s) =π−2

Z ∞

rm

sdr

r

q

r2₍₁₋V(r)

E )−s2

What isV(r) for our central force off =k/r3_{? It’s found from−dV /dr=f.}

Plug this in to Θ and we have

Θ(s) =π−2

Z ∞

rm

sdr

rqr2_−(s2₊ k

2E)

Before taking this integral, I’d like to put it in a better form. If we look at the energy of the incoming particle,

E= 1 2mr

2

mθ˙

2₊ k

2r2 m = s 2_E r2 m + k 2r2 m

where from Goldstein page 113,

˙ θ2= 2s

2_E

mr4

m

We can solve fors2+₂k_E, the term in Θ,

r2_m=s2+ k 2E Now we are in a better position to integrate,

Θ(s) =π−2

Z ∞

rm

sdr

rpr2_−r2

m

=π−2s[ 1 rm

cos−1rm r ∞ rm

] =π−2s 1 rm

(π

2) =π(1− s

q

s2₊ k

2E

)

Goldstein gave usx= Θ/π, so now we have an expression forxin terms of s, lets solve fors

x=Θ π = 1−

s

q

s2₊ k

2E

s2= (s2+ k

2E)(1−x)

2

→ s2=

k

2E(1−x)

2

1−(1−x)2

s=

r

k 2E

(1−x)

p

x(2−x)

Now that we have swe need only ds/dΘ to find the cross section. Solving dΘ/dsand then taking the inverse,

dΘ

ds =πs(− 1 2(s

2₊ k

2E)

−3

2)2s+ π

q

s2₊ k

2E

dΘ ds =

−πs2_+π(s2₊ k

2E)

(s2₊ k

2E)

3 2

=

πk

2E

(s2₊ k

2E)

3 2

ds dΘ =

2E(s2+₂k_E)32

πk Putting everything in terms ofx,

s2+ k 2E =

k 2E

(1−x)2

x(2−x)+ k 2E =

k 2E

1 x(2−x)

So now,

σ(Θ) = s sin Θ

ds dΘ

=

q

k

2E

(1−x) √

x(2−x)

sinπx

2E(s2₊ k

2E)

3 2

πk =

q

k

2E

(1−x) √

x(2−x)

sinπx

2E(₂k_Ex(21_−x))32

πk =

And this most beautiful expression becomes..

σ(Θ) = 1 sinπx

1 π(

k 2E)

1 2(2E

k )( k 2E)

3

2 1−x

p

x(2−x) 1

(x(2−x))32

After a bit more algebra...

σ(Θ) = k 2E

1 π

1 sinπx

1−x (x(2−x))2

And since we knowdΘ =πdx,

σ(Θ)dΘ = k 2E

(1−x)dx x2_(2−x)2_sinπx

3.32 A central force potential frequently encountered in nuclear physics is the rectangular well, defined by the potential

V = 0 r > a

V =−V0 r≤a

Show that the scattering produced by such a potential in classical mechanics is identical with the refraction of light rays by a sphere of radius a and relative index of refraction

n=

r

E+V0

E

This equivalence demonstrates why it was possible to explain refraction phe-nomena both by Huygen’s waves and by Newton’s mechanical corpuscles. Show also that the differential cross section is

σ(Θ) = n

2_a2

4 cosΘ₂

(ncosΘ₂ −1)(n−cosΘ₂)

(1 +n2_−2ncosΘ 2)2

Answer:

Ignoring the first part of the problem, and just solving for the differential cross section,

σ(Θ) = sds sin ΘdΘ

If the scattering is the same as light refracted from a sphere, then putting our total angle scattered, Θ, in terms of the angle of incidence and transmission,

Θ = 2(θ1−θ2)

This is because the light is refracted from its horizontal direction twice, after hitting the sphere and leaving the sphere. Whereθ1−θ2 is the angle south of

east for one refraction.

We know sinθ1=s/a and using Snell’s law, we know

n= sinθ1 sinθ2

→sinθ2=

s na

Expressing Θ in terms of justsandawe have

Θ = 2(arcsins

a−arcsin s na)

Now the plan is, to solve for s2 _{and then ds}2_{/dΘ and solve for the cross}

section via

σ= sds sin ΘdΘ=

1 2 sin Θ

ds2 dΘ =

1 4 sinΘ₂ cosΘ₂

ds2 dΘ

Here goes. Solve for sinΘ₂ and cosΘ₂ in terms ofs

sinΘ

2 = sin(arcsin s

a−arcsin s

na) = sin arcsin s

acos arcsin s

na−cos arcsin s

asin arcsin s na

This is

= s

acos(arccos

r

1− s 2

n2_a2)−cos(arccos(

r

1−s 2

a2)

s na

Using arcsinx= arccos√1−x2_{and sin(a−b) = sinacosb−cosasinb. Now}

we have

sinΘ 2 =

s na2(

p

n2_a2_−s2₎₋p_a2_−s2₎

cosΘ 2 =

1 na2(

p

a2_−s2p_n2_a2_−s2_+s2₎

Using cos(a−b) = cosacosb+ sinasinb. Still solving fors2_{in terms of cos}

and sin’s we proceed

sin2Θ 2 =

s2

n2_a4(n

2_a2_−s2₋₂p

n2_a2_−s2p_a2_−s2_+a2_−s2₎

This is

sin2 s

2

n2_a2(n

2_{+ 1)−} 2s 4

n2_a4 −

2s2 n2_a4

p

n2_a2_−s2p_a2_−s2

Note that

p

n2_a2_−s2p_a2_−s2_=na2_cosΘ

2 −s

2

So we have

sin2Θ 2 =

s2

n2_a2(n 2_{+ 1}

−2s 2

a2 −2ncos

Θ 2 +

2s2

a2 ) =

s2

n2_a2(1 +n 2

−2ncosΘ 2)

Solving fors2

s2= n

2_a2_sin2 Θ 2

1 +n2_−2ncosΘ 2

Glad that that mess is over with, we can now do some calculus. I’m going to letq2 _{equal the denominator squared. Also to save space, lets say} Θ

2 =Q. I

like using the letterq.

ds2 dΘ =

a2sinQn2

q2 [cosQ(1−2ncosQ+n

2_)−nsin2_Q]

ds2

dΘ = n2_a2

q2 sinQ[cosQ−2ncos

2_Q+n2_{cosQ−n(1−cos}2_Q)]

Expand and collect

ds2

dΘ = n2_a2

q2 sinQ[−ncos

2_{Q+ cosQ+n}2_cosQ−n]

Group it up

ds2

dΘ = n2_a2

q2 sinQ(ncosQ−1)(n−cosQ)

ds2

dΘ =

n2_a2_sinΘ 2(ncos

Θ

2 −1)(n−cos Θ

2)

(1−2ncosΘ₂ +n2₎2

Using our plan from above,

σ= 1

4 sinΘ 2 cos

Θ 2

ds2

dΘ = 1 4 sinΘ

2 cos Θ

2

n2_a2_sinΘ 2(ncos

Θ

2 −1)(n−cos Θ 2)

(1−2ncosΘ 2 +n

2₎2

We obtain

σ(Θ) = 1 4 cosΘ₂

n2_a2_(ncosΘ

2 −1)(n−cos Θ

2)

(1−2ncosΘ₂ +n2₎2

The total cross section involves an algebraic intensive integral. The total cross section is given by

σT = 2π

Z Θmax

0

σ(Θ) sin ΘdΘ

To find Θmax we look for when the cross section becomes zero. When

(ncosΘ₂ −1) is zero, we’ll have Θmax. Ifs > a, its as if the incoming particle

misses the ‘sphere’. Ats=awe have maximum Θ. So using Θmax= 2 arccos_n1,

we will find it easier to plug inx= cosΘ₂ as a substitution, to simplify our in-tegral.

σT =π

Z 1

1

n

a2n2(nx−1)(n−x) (1−2nx+n2₎22dx

where

dx=−1

2sin Θ

2dΘ cos

Θmax

2 =

1 n

The half angle formula, sin Θ = 2 sinΘ₂ cosΘ₂ was used on the sin Θ, the negative sign switched the direction of integration, and the factor of 2 had to be thrown in to make thedxsubstitution.

This integral is still hard to manage, so make another substitution, this time, letqequal the term in the denominator.

q= 1−2nx+n2 → dq=−2ndx

The algebra must be done carefully here. Making a partial substitution to see where to go:

qmin= 1−2 +n2=n2−1 qmax=n2−2n+ 1 = (n−1)2

σT =

Z (n−1)2

n2₋₁

2πa2_n2_{(nx−1)(n−x)}

q2

dq

−2n =πa

2Z (n−1)2

n2₋₁

−n(nx−1)(n−x)

Expandingq2to see what it gives so we can put the numerator in the above integral in terms ofq2 we see

q2=n4+ 1 + 2n2−4n3x−4nx+ 4n2x2

Expanding the numerator

−n(nx−1)(n−x) =−n3x−nx+n2x2+n2

If we takeq2_{and subtract an}4_{, subtract a 1, add a 2n}2_{and divide the whole}

thing by 4 we’ll get the above numerator. That is:

q2−n4+ 2n2−1

4 =

q2−(n2−1)2

4 =−n(nx−1)(n−x) Now, our integral is

σT =πa2

Z (n−1)2

n2₋₁

q2_−(n2₋₁₎2

4q2 dq

This is finally an integral that can be done by hand

σT =

πa2

4

Z

1−(n 2₋₁₎2

q2 dq=

πa2

4 (z+

(n2₋₁₎2

z

(n−1)2

n2₋1

)

After working out the few steps of algebra,

πa2

4

4n2_{−8n+ 4}

n2_{−2n+ 1} =πa 2

The total cross section is